A student decides to move a box of books into her dormitory room by pulling on a

Question

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 160 N at an angle of 34 degrees above the horizontal. The box has a mass of 25 kg, and the coefficient of friction between box and floor is 0.271. Find the acceleration of the box. I have solved this part of the problem, getting 3.612 for the acceleration. I have not sure how to approach part b of this problem:
The student now starts moving the box up a 13 degree incline, keeping her 160 N force directed at 34 degrees above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?

Explanation / Answer

a) First of all, resolve the 160 N force into its x and y components to make solving this easier. In the x-direction, F = (160 N)cos34 = 132.64 N In the y-direction, F = (160 N)sin34 = 89.47 N The forces in the y-direction are gravity, the y-component of the applied force, and the normal force. The only unknown here is the normal force since you know the force due to gravity (9.8 m/s^2 * 25 kg = 245N) and the y-component of the applied force. Net force in the y = 0 --> 89.47 N - 245 N + F(normal) = 0 --> F(normal) = 155.53 N so now you have the normal force which you need to find the force of friction. Remember, F(friction) = (coefficient of friction)*(normal force) --> F(friction) = (0.271)(155.53 N) = 42.14 N Remember that friction always opposes the motion of an object. Therefore, net force in the x direction = Force applied in the x direction - Force of friction = 132.64 N - 42.14 N = 90.49 N Therefore, the net force acting on the box is 90.49 N in the x direction. You can now find the acceleration of the box using F = ma --> 90.49 N = (25 kg)a --> a = 3.619 m/s^2. b) This time resolve all the forces into forces parallel to the incline and forces perpendicular to the incline. I'll call x the parallel direction and y the perpendicular (not to be confused with horizontal and vertical). Applied force in the x direction = 132.64 N Applied force in the y direction = 89.47 N Force of gravity in the x = (9.8)(25 kg)(sin 13) = 53.42 N Force of gravity in the y = (9.8)(25 kg)(cos 13) = 231.41 N Again, find the normal force, noting that the net force perpendicular to the ramp is 0 since the box is not accelerating in that direction: Net force y = 0 --> 89.47 N - 231.41 N + F(normal) = 0 --> F(normal) = 141.9 N The force of friction = (141.94 N)(0.271) = 38.46 N Ok, so now you have all the forces in the x direction. Net force in the x = 132.64 N - 38.46 N - 53.42 N = 40.76 N F = ma --> 40.76 N = (25 kg)a --> a = 1.63 m/s^2 UP the ramp.

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